Rotational Motion • Rotation (rigid body) versus translation (point particle) • Rotation concepts and variables • Rotational kinematic quantities Angular position and displacement Angular velocity Angular acceleration • Rotation kinematics formulas for constant angular acceleration 1 “Radian” “radian” : more convenient unit for angle than degree Definition: 360o 180o • 2π radians = 360 degree 1 radian = = = 57.3o 2π π s arc length ≡ s = r θ (in radians) θ ≡ rad r θ (in rad) s =×2π rr =θ 2π r s θ’ θ Example: r = 10 cm, θ = 100 radians Æ s = 1000 cm = 10 m. 1 Rigid body Rigid body: A “rigid” object, for which the position of each point relative to all other points in the body does not change. Example: Solid: Rigid body Liquid: Not rigid body Rigid body can still have translational and rotational motion. Angular position of rotating rigid body • By convention, θ is measured CCW from the x-axis • It keeps increasing past 2π, can be y negative, etc. • Each point of the body moves around θ the axis in a circle with some specific Reference x line rotates radius with body rotation axis “o” rigid body fixed to body parallel to z-axis 2 Angular displacement of rotating rigid body y θ Reference x line rota tes with body Angular displacement: rotation axis “o” rigid body • Net change in the angular coordinate fixed to body parallel to z-axis Δθ ≡ θfinal − θinital (an angle in rad.) Arc length: Δs • Measures distance covered by a point as it moves Reference line rotating with body y through Δθ (constant r) Δs = r Δθ Δs ≡ rΔθ (a distance along a circular arc) θf r r θo x Rigid body rotation: angular & tangential velocity Angular velocity ω: For any point, r is the perpendicular • Rate of change of the angular displacement distance to the rotation axis Δθ Δθ dθ ωave ≡ ωinst ≡ Lim ≡ Δt Δt → 0 Δt dt vT • Units: radians/sec. Positive in Counter-Clock-Wise sense • Frequency f = # of complete revolutions/unit time r • f = 1/T T = period (time for 1 complete revolution θ = ωΔτ x ω = 2πf = 2π/T f = ω/2π Tangential velocity vT: • Rate at which a point sweeps out arc length along circular path ΔΔs θ Æ Δ=Δs r θ Æ = r vrT = ω ΔΔtt 3 iClicker Quiz 1.1. The period of a rotating wheel is 12.57 seconds. The radius of the wheel is 3 meters. It’s angular speed is closest to: A. 79 rpm B. 0.5 rad/s C. 2.0 rad/s D. .08 rev/s E. 6.28 rev/s 1.2. A point on the rim of the same wheel has a tangential speed closest to: A. 12.57 rev/s B. 0.8 rev/s C. 0.24 m/s Δs ≡ rΔθ D. 1.5 m/s E. 6.28 m/s vT = ωr ω = 2πf = 2π/T Rigid body rotation: angular acceleration Angular acceleration α: Δω Δωωd • Rate of change of the angular velocity ααave ≡≡ inst Lim = ΔtdtΔ→t 0 Δt • Units: rad/s2 • CCW considered positive • for CONSTANT α: ωf = ω0 + αΔt 4 1D and Angular Kinematics Equations (Same mathematical forms) 1D motion with Angular motion with constant acceleration a constant angular acceleration α x(t), v(t), a(t) variables θ(t), ω(t), α(t) dx dv dθ dω v = a = Definitions ω = α = dt dt dt dt vf (t) = v0 + at Kinematic ωf (t) = ω0 + αt Equations x (t) = x + v t + 1 at 2 θ (t) = θ + ω t + 1 αt2 f 0 0 2 f 0 0 2 v2 (t) = v2 + 2a[x − x ] 2 2 f 0 f 0 ωf (t) = ω0 + 2α[θf − θ0 ] Rotational variables are vectors, having direction The angular displacement, speed, and acceleration ( θ, ω, α ) are vectors with direction. The directions are given by the right-hand rule: Fingers of right hand curl along the angular direction (See Fig.) Then, the direction of thumb is thdihe direct ion o fhf the angu lar quantity. 5 Example: A grindstone is rotating with constant angular acceleration about a fixed axis in space. Initial conditions 2 at t = 0: α = 0.35 rad/s ω0 = - 4.6 rad/s When is Δθ = 0 again in addition to t=0? Positive directions: right hand rule Example: Wheel rotating and accelerating At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration has an angular velocity of 2.0 rad/s. Two seconds later it has turned through 5.0 complete revolutions. Find the angular acceleration of this wheel? ωf (t) = ω0 + αt θ (t) = θ + ω t + 1 αt2 f 0 0 2 2 2 ωf (t) = ω0 + 2α[θf − θ0 ] 6 Rigid body rotation: radial and tangential acceleration Centripetal (radial) acceleration ac or ar • Radial acceleration component, points toward rotation axis v2 T 2 Fma= arr == ωω (use v = r ) rr a r T T vT r ac ω,α Tangential acceleration aT: x • Tangential acceleration component • Proportional to angular acceleration α and also to radius r • Units: length / time 2 aT = rα Fmatangential = T Rotation variables: angular vs. linear s = rΔθ vrT = ω arT = α v 2 a = T =rω 2 r r 7 A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s angular velocity is A. half the ladybug’s. B. the same as the ladybug’s. C. twice the ladybug’s. D. impossible to determine A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s velocity is A. half the ladybug’s. B. the same as the ladybug’s. C. twice the ladybug’s. D. impossible to determine 8 Rotational Dynamics We want something like “F=ma” for rotational motion….. • Moment of inertia – rotational analog of mass • Torque – rotational analog of force 17 Something like mass for rotational motion: Moment of Inertia, I G L Kinetic energy of ladybug and gentlemanbug 11 1 1 Kmvmvmr=+22 =()ωω 2 + mr () 2 22LL GG 2 L L 2 G G 1122 22 1 2 2 2 1 2 =+mrLLω mr GGωωω =+=( mr LL mr GG ) I 22 2 2 22 I =+mrL LGG mr 1 Generally, Imrmrmr=+++222... Kinetic energy: K = Iω 2 11 2 2 33 2 9 Example: Find moment of inertia for a crossed dumbbell •Four identical balls as shown: m m = 2 kg •Connected by massless rods: 2 d d length d = 1 m. mmdd A BC d Rotational inertia I depends on axis chosen m A) Choose rotation axis perpendicular to figure through point “A” B) Now c hoose ax is perpen dicul ar t o fi gure th rough poi nt “B” C) Let rotation axis pass through points “B” and “C” Calculation of Moment of inertia for continuous mass distributions requires “Integration, a kind of calculus”. We will just use the result. 10 Moments of Inertia of Various Rigid Objects Now we want to define “torque, τ”, so that “τ = I α”. F Newton’s Law along tangential direction θ F T m FTT= ma= m r α r Multiplying “r”, so that we have “I” on right side rp axis 2 rFT == m r α Iα So, let’s define torque as τ ≡ rFT Then we got τ = Iα Since FFT = sinθ and rrp = sinθ τ ==rFTp rFsinθ = r F (could be positive or negative) 11 F τ ==rFTp rFsinθ = r F θ F T m If r = 0, torque is zero. r rp axis If theta = 0 or 180 degree, the torque is zero. For multiple forces τ net =+++τττ123... τ net = Iα m1 m2 m1=100 kg adult, m2=10 kg baby. Distance to fulcrum point is 1 m and 11 m respectively. The seesaw s tar ts at hor izon tal positi on f rom rest. Which direction will it rotates? (a) Counter-Clockwise (b) Clockwise (c) No rotation (d) Not enough information Example: Find the net torque, moment of inertia, and initial angular acceleration. Choose axis of rotation through fulcrum point. 12 Example: second law for rotation PP10606-49*: When she is launched from a springboard, a diver's angular speed about her center of mass changes from zero to 6.20 rad/s in 220 ms. Her rotational inertia about her center of mass is constant at 12.0 kg·m2. During the launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external torque on her from the board? G G τnet = Itot α 5 N tangential force is applied at the edge of a uniform disk of radius 2 m and mass of 8 kg. Find angular acceleration. F=5 N Axis of rotation 1 Formula: I = MR2 2 13 5 N tangential force is applied at 1 m from the center of a uniform disk of radius 2 m and mass of 8 kg. Find angular acceleration. FF5=5N N Axis of rotation 1 Formula: I = MR2 2 5 N force is applied at 1 m from the center of a uniform disk of radius 2 m and mass of 8 kg. Find torque. 45 degree F=5 N Axis of rotation 14 Torque on extended object by gravitational force Æ Assume that the total gravitational force effectively acts at the center of mass.